A cube is placed at the origin as shown in the figure. The electric field in the region is given as $\vec E = (200 x^2\hat i)\frac{N}{C}$. The flux coming out of the cube is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (2) → $3200\frac{Nm^2}{C}$
Electric field: $\vec{E}=(200x^2\hat{\imath})\ \mathrm{N/C}$
Only faces normal to $\hat{\imath}$ contribute. At $x=2$ (outward normal $+\hat{\imath}$):
$\Phi_{x=2}=\iint_{0}^{2}\iint_{0}^{2}E_x(2)\,dy\,dz=E_x(2)\cdot(2\cdot2)=200(2)^2\cdot4=3200\ \mathrm{N\,m^2/C}$
At $x=0$ (outward normal $-\hat{\imath}$): $E_x(0)=0\Rightarrow \Phi_{x=0}=0$
Total flux: $\Phi=\;3200\ \mathrm{N\,m^2/C}$